Select the single best answer. Determine the following type of reaction: CH3―CH2―CH(Br)―CH3 CH3―CH═CH―CH3 + NaBr + H2O
1 answer:
Answer:
Deshydrohalogenation
Explanation:
You are not providing options to answer, however, this can be answered without options.
Now, in the reaction we can see that we have an atom of Bromine in carbon 2, and in the product appears as NaBr. This means that the Br was substracted by elimination. It's an elimination because the final product do not have a substituent where the bromine was, (Like another nucleophyle such OH or another halide). If you look closely the final product, we can see that one hydrogen in carbon 3, is no longer there. So this electrophyle was also substracted, in this case, by a base (Such NaOH), so in this case, it's ocurring an elimination reaction via E2 (One step, bimolecular). So, as the final product has been substracted the nucleophyle and electrophyle, this treaction is a deshydrohalogenation (an atom of hydrogen and a halide were substracted). The mechanism of this, you can see it in the picture.
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Answer:
The dots were not properly located and arrows are not used in Lewis structures
Explanation:
If we intend to write a Lewis structure for a compound, that lewis structure must consist of only dots. These dots actually show the valence electrons on the outermost shell of the molecule.
We do not involve arrows when writing dot electron structures for compounds. The valence electrons of magnesium ought not to be written together because they are not a lone pair, rather they are two unpaired electrons. The use of an arrow suggests a coordinate covalent bond which is not the case here.
The correct lewis structure for MgCl2 is shown in the image attached to this answer.
